The topic of this problem is Kirchhoff's Current Law. In this problem, we want to determine I1 and I2. So we have a circuit that's drawn, and it has a voltage source in it. It has four resistors and some currents that are already known. The unknown currents are I1 and I2. So to find these currents, we're going to use Kirchhoff's Current Law, and we're going to sum the currents into the nodes. So the first thing we have to do when we solve problems using Kirchhoff's Current Law is we have to identify the nodes in the problem. So we have a common node at the bottom. We have a node on the left hand side of the circuit, and we have a node on the right hand side of the circuit. So there's three nodes associated with this circuit. So if we want to use Kirchhoff's Current Law, we can sum the current into any one of those nodes and see if we can solve for quantities that we have of interest. So let's start with the node on the left hand side of the circuit. So we want to sum the currents into the left hand node at the top. And so if we do that, we see first of all, that we have I1 flowing into that node. So it's I1 and then we have three amps flowing out through the left most resistor. So it's a -3 mA, and we also have I2, which is flowing out of that part of the circuit as well. And so that node has the equation as shown above, so that is node 1 and we'll assume that is equation 1. If we go to the other side of the circuit and look at node 2, and we do the same thing. We sum the currents into that node, then we see that I1 is flowing out, so it's -I1, and we have 12mA that's flowing up into node 2 through the voltage source. So it's +12 mA. And we have 4 mA flowing down through and out of node 2 through the resistor on the rightmost side of the circuit, so it's -4mA. And that's equal to 0. So, we can see that we can solve for I1, and then alpha mean for I2. So, I1 in this case is going to be 8 mA, and if I1 is 8 mA, then I2 using equation 1 turns out to be 5 mA.
